Mauritius
Two concentric circles have radii ^@a^@ and ^@b^@ ^@(a>b)^@. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
^@ 2 \sqrt { a^2 - b^2 } ^@
- Let ^@O^@ be the common center of the two circles and ^@AB^@ be the chord of the larger circle which touches the smaller circle at ^@C^@.
As the radius of the larger circle is ^@a^@ and that of the smaller circle is ^@b^@, we have ^@OA = a ^@ and ^@OC = b^@.
The situation given can be represented by the image below.
- ^@AB^@ is the tangent to the circle with radius ^@b^@.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
So, ^@OC \perp AB. ^@
^@AB^@ is the chord of the circle with radius ^@a^@.
The perpendicular drawn from the center of the circle to a chord bisects the chord.
^@ \therefore OC ^@ bisects ^@AB \implies C ^@ is the mid-point of ^@AB^@. - Using Pythagoras' theorem in the right- angled triangle ^@ACO^@, we have @^ \begin{aligned} & OA^2 = OC^2 + AC^2 && \\ \implies & AC = \sqrt { OA^2 - OC^2 } = \sqrt { a^2 - b^2 } \end{aligned} @^ ^@C^@ is the mid-point of ^@AB.^@ So, ^@AB = 2AC = 2 \sqrt { a^2 - b^2 }. ^@
- Thus, the length of the chord of the larger circle which touches the smaller circle is ^@ 2 \sqrt { a^2 - b^2 } ^@.
